Download Physics 207: Lecture 2 Notes

Physics 207, Lecture 11, Oct. 11
Agenda: Chapter 7, finish, Chapter 8, Potential Energy
• Work-Energy Theorem
• Work and Friction
• Power
•
 P = dW / dt = F · v (a vector product!)
Potential Energy
Conservative Forces and Potential Energy (W = -DU)
Non-conservative Forces
Generalized Work Energy Theorem
Assignment: For Monday read Chapter 9
 WebAssign Problem Set 4 due Tuesday next week
Physics 207: Lecture 11, Pg 1
See text: 7-4
Work Kinetic-Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
Wnet  DK
 K 2  K1
(final – initial)
1
1
2
2
 mv 2  mv1
2
2
Physics 207: Lecture 11, Pg 2
Lecture 11, Exercise 1
Kinetic Energy (= ½ mv2)

To practice your pitching you use two baseballs.
The first time you throw a slow curve and clock the
speed at 50 mph (~25 m/s). The second time you
go with high heat and the radar gun clocks the pitch
at 100 mph. What is the ratio of the kinetic energy
of the fast ball versus the curve ball ?
(A) 1/4
(B) 1/2
(C) 1
(D) 2
(E) 4
Physics 207: Lecture 11, Pg 3
Example: Work Kinetic-Energy Theorem
•
How much will the spring compress (i.e. Dx) to bring
the object to a stop (i.e., v = 0 ) if the object is moving
initially at a constant velocity (vo) on frictionless surface
as shown below ?
to vo
Notice that the
spring force is
opposite to the
displacemant.
F
m
spring at an equilibrium position
For the mass m,
work is negative
Dx
V=0
t
m
For the spring, work
is positive
spring compressed
Physics 207: Lecture 11, Pg 4
Another Example
A mass m starts at rest and is attached to a spring which is expanded a
distance L / 2 to the left of its equilibrium position and then is allow to
move a distance L to the right where the mass is once again at rest.
Question: How much work was done on the box? (Notice DK = 0)
xf
v=
Fon m by spring
Wbox  F ( x ) dx
0
xi
L
m
Wbox   kx dx
Fon spring by m
0


equilibrium position
-L /2
x=L
 - 12 kx 2 |L  - kL2 ???
0
This doesn’t make sense!
x=0
V=0
Wbox
L/2
m
Wbox
   kx dx
L / 2
spring compressed
Wbox  - 12 k ( L / 2) 2  12 k ( L / 2) 2  0
Physics 207: Lecture 11, Pg 5
Lecture 11, Exercise 2
Work & Friction



Two blocks having mass m1 and m2 where m1 > m2.
They are sliding on a frictionless floor and have the same
kinetic energy when they encounter a long rough stretch
(i.e. m > 0) which slows them down to a stop.
Which one will go farther before stopping?
Hint: How much work does friction do on each block ?
(A) m1
(B) m2
m1
m2
(C) They will go the same distance
v1
v2
Physics 207: Lecture 11, Pg 6
Lecture 11, Exercise 3
Work & Friction

You like to drive home fast, slam on your brakes at the start of
the driveway, and screech to a stop “laying rubber” all the
way. It’s particularly fun when your mother is in the car with
you. You practice this trick driving at 20 mph and with some
groceries in your car with the same mass as your mother. You
find that you only travel half way up the driveway. Thus when
your mom joins you in the car, you try it driving twice as fast.
How far will you go this time ?
A. The same distance. Not so exciting.
B.  2 times as far (only ~7/10 of the way up the driveway)
C. Twice as far, right to the door. Whoopee!
D. Four times as far crashing into the house. (Oops.)
Physics 207: Lecture 11, Pg 7
Work & Power:





Two cars go up a hill, a Corvette and a ordinary
Chevy Malibu. Both cars have the same mass.
Assuming identical friction, both engines do the same
amount of work to get up the hill.
Are the cars essentially the same ?
NO. The Corvette gets up the hill quicker
It has a more powerful engine.
Physics 207: Lecture 11, Pg 8
Work & Power:



Power is the rate at which work is done.
Average Power is,
W
P
Instantaneous Power is,
Dt
dW  
P
 F V
dt
Physics 207: Lecture 11, Pg 9
Lecture 11, Exercise 4
Work & Power

Starting from rest, a car drives up a hill at constant
acceleration and then suddenly stops at the top. The
instantaneous power delivered by the engine during
this drive looks like which of the following,

(A)


(B)
(C)
time
time
time
Physics 207: Lecture 11, Pg 10
Lecture 11, Exercise 5
Power for Circular Motion

I swing a sling shot over my head. The tension in the rope
keeps the shot moving in a circle. How much power must
be provided by me, through the rope tension, to keep the
shot in circular motion ?
Note that: Rope Length = 1m
Shot Mass = 1 kg
Angular frequency = 2 rad / sec
(A) 16 J/s
(B) 8 J/s
(C) 4 J/s
v
(D) 0
Physics 207: Lecture 11, Pg 11





Lecture 11, Exercise 5
Power for Circular Motion
Note that the string expends no power ( because it does
no work).
By the work / kinetic energy theorem, work done equals
change in kinetic energy.
K = 1/2 mv2, thus since v doesn’t change, neither does
K.
A force perpendicular to the direction of motion does not
change speed, |v|, and so does no work.
Answer is (D)
v
T
Physics 207: Lecture 11, Pg 12
Chapter 8, Potential Energy


What is “Potential Energy” ?
It is a way of effecting energy transfer in a system so that
it can be “recovered” (i.e. transferred out) at a later time or
place.
Example: Throwing a ball up a height h above the ground.
No Velocity at time 2
but DK = Kf - Ki= -½ m v2
Velocity v up
at time 1
Velocity v down
at time 3
Energy is conserved and so work must have been done
Physics 207: Lecture 11, Pg 13
We can identify this work:

Consider a ball moving up to height h above the ground.
(from time 1 to time 2)
What work is done in this process ?
(Work done by the earth on the ball)
mg
W = F  Dx
h
W = -mgh and W = - mgh = -½ m v2
mg
Physics 207: Lecture 11, Pg 14
Potential Energy

Now consider the ball starting a height h above the
ground and falling (from time 2 to time 3)
h
Before the ball falls it has the potential to do
an amount of work mgh.
We say the ball has a potential energy of
U = mgh.
By falling the ball loses its potential energy,
work is done on the ball, and it gains some
kinetic energy,
W = DK= ½ m v2 = -DU = -(Ufinal - Uinit) = mgh
Notice that potential is measured as a difference with respect
to an arbitrary reference (here it is the ground).
Physics 207: Lecture 11, Pg 15
Potential Energy, Energy Transfer and Path
A ball of mass m, initially at rest, is released and follows three
difference paths. All surfaces are frictionless
1. The ball is dropped
2. The ball slides down a straight incline
3. The ball slides down a curved incline
After traveling a vertical distance h, how do the three speeds compare?

1
2
3
h
(A) 1 > 2 > 3
(B) 3 > 2 > 1
(C) 3 = 2 = 1 (D) Can’t tell
Physics 207: Lecture 11, Pg 16
Lecture 11, Exercise 7
Work Done by Gravity


An frictionless track is at an angle of 30° with respect
to the horizontal. A cart (mass 1 kg) is released. It
slides 1 meter downwards along the track bounces
and then slides upwards to its original position.
How much total work is done by gravity on the cart
when it reaches its original position? (g = 10 m/s2)
30°
(A) 5 J
(B) 10 J
(C) 20 J
(D) 0 J
Physics 207: Lecture 11, Pg 17
Work Done (by the person) Against Gravity


Consider lifting a box onto the tail gate of a truck.
Condition: Box is rising at constant velocity
N = mg
m
h
mg
The work required for this task is,
W=F·d=Nh
W = mgh
Note: Holding the box level involves NO work
Physics 207: Lecture 11, Pg 18
Work Done Against Gravity

Now use a ramp, of length L, to help you with the task.
Is less work needed to get the box into the truck?
(It’s “easier” to lift the box) and it moves at constant
velocity.
N
-mg sin q
h
mg
The work required for this task is,
W = F · d = -mg sin q (L)
mg sin q
and L sin q = h
W = mgh  Work is identical but force is not.
Physics 207: Lecture 11, Pg 19
Important Definitions

Conservative Forces - Forces for which the work
done does not depend on the path taken, but only the
initial and final position (no loss).

Potential Energy - describes the amount of work that
can potentially be done by one object on another
under the influence of a conservative force
 W = -DU
Only differences in potential energy matter.
Physics 207: Lecture 11, Pg 20
Potential Energy

For any conservative force F we can define a
potential energy function U in the following way:
W =

 F.dr = - DU
The work done by a conservative force is equal and
opposite to the change in the potential energy function.
r2

U2
This can be written as:
 F.dr
r2
DU = U2 - U1 = - W = -
r1
r1
U1
Physics 207: Lecture 11, Pg 21
A Conservative Force Example: Hooke’s Law Spring

For a spring we know that Fx = -kx.
F(x)
x1
x2
x
Equilibrium position
-kx
F = - k x1
F = - k x2
Physics 207: Lecture 11, Pg 22
Spring...


The work done by the spring Ws during a displacement
from x1 to x2 is the area under the F(x) vs x plot
between x1 and x2.
A spring under compression or tension has stored
potential energy
F(x)
x1
x2
x
Ws
-kx
Active Figure
Physics 207: Lecture 11, Pg 23
See text: 8-4
Conservation of Energy

If only conservative forces are present, the total energy
(sum of potential and kinetic energies) of a system is
conserved.
E=K+U
E = K + U is constant !!!

Both K and U can change, but E = K + U remains constant.
E is called “mechanical energy”
Physics 207: Lecture 11, Pg 24
A Non-Conservative Force
Friction


Looking down on an air-hockey table with no air
flowing (m > 0).
Now compare two paths in which the puck starts out
with the same speed (K1 = K2) .
Path 2
Path 1
Physics 207: Lecture 11, Pg 25
A Non-Conservative Force
Path 2
Path 1
Since path2 distance >path1 distance the puck will be
traveling slower at the end of path 2.
Work done by a non-conservative force irreversibly
removes energy out of the “system”.
Here WNC = Efinal - Einitial < 0
Physics 207: Lecture 11, Pg 26
Lecture 11, Exercise 8
Work/Energy for Non-Conservative Forces

The air track is once again at an angle of 30° with
respect to horizontal. The cart (with mass 1 kg) is
released 1 meter from the bottom and hits the bumper
at a speed, v1. This time the vacuum/ air generator
breaks half-way through and the air stops. The cart only
bounces up half as high as where it started.

How much work did friction do on the cart ?(g=10 m/s2)
30°
(A) 2.5 J (B) 5 J (C) 10 J (D) –2.5 J (E) –5 J (F) –10 J
Physics 207: Lecture 11, Pg 27
Another example of a conservative system:
The simple pendulum.

Suppose we release a mass m from rest a distance
h1 above its lowest possible point.
 What is the maximum speed of the mass and
where does this happen ?
 To what height h2 does it rise on the other side ?
m
h1
h2
v
Physics 207: Lecture 11, Pg 28
Example: The simple pendulum.
 What is the maximum speed of the mass and
where does this happen ?
E = K + U = constant and so K is maximum when
U is a minimum.
y
y=h1
y=0
Physics 207: Lecture 11, Pg 29
Example: The simple pendulum.
What is the maximum speed of the mass and
where does this happen ?
E = K + U = constant and so K is maximum when
U is a minimum
E = mgh1 at top
E = mgh1 = ½ mv2 at bottom of the swing
y
y=h1
y=0
h1
v
Physics 207: Lecture 11, Pg 30
Example: The simple pendulum.
To what height h2 does it rise on the other side?
E = K + U = constant and so when U is maximum
again (when K = 0) it will be at its highest point.
E = mgh1 = mgh2 or h1 = h2
y
y=h1=h2
y=0
Physics 207: Lecture 11, Pg 31
Lecture 11, Exercise 9
The Loop the Loop … again


To complete the loop the loop, how high do we
have to let the release the car?
Condition for completing the loop the loop: Circular
motion at the top of the loop (ac = v2 / R)
Car has mass m
h?
Recall that “g” is the source
of this acceleration and N
goes to zero. that to
avoid death, the minimum
speed at the top is v  gR
R
(A) 2R
(B) 3R
(C) 5/2 R
(D) 23/2 R
Physics 207: Lecture 11, Pg 32
See text: 8.5
Non-conservative Forces :

If the work done does not depend on the path taken,
the force involved is said to be conservative.

If the work done does depend on the path taken, the
force involved is said to be non-conservative.

An example of a non-conservative force is friction:

Pushing a box across the floor, the amount of work that
is done by friction depends on the path taken.
 Work done is proportional to the length of the path !
Physics 207: Lecture 11, Pg 33
Generalized Work Energy Theorem:

Suppose FNET = FC + FNC (sum of conservative and nonconservative forces).

The total work done is: WTOT = WC + WNC

The Work Kinetic-Energy theorem says that: WTOT = DK.
 WTOT = WC + WNC = DK

WNC = DK - WC

But WC = -DU
So
WNC = DK + DU = DE
Physics 207: Lecture 11, Pg 34
Physics 207, Lecture 11, Recap
Agenda: Chapter 7, finish, Chapter 8, Potential Energy
• Work-Energy Theorem
• Work and Friction
• Power
•
 P = dW / dt = F · v (a vector product!)
Potential Energy
Conservative Forces and Potential Energy (W = -DU)
Non-conservative Forces
Generalized Work Energy Theorem
Assignment: For Monday read Chapter 9
 WebAssign Problem Set 4 due Tuesday next week
Physics 207: Lecture 11, Pg 35